On efficient abstract method for the study of Cauchy
problem for a second order differential equation set
on a singular cylindrical domain
Belkacem Chaouchi
1*
Abstract
In this work, we present an abstract approach for the study of a mixed initial value problem set in cylindrical domain with a cusp base. MaximalLpregularity results for the strict solution are established.
Keywords
Fractional powers of linear operators; analytic semigroup, Abstract differential equation, Cuspidal point.
AMS Subject Classification
34G10, 34K10, 12H2O, 44A45.
1Lab. de l’Energie et des Syst `emes Intelligents, Khemis Miliana University, Algeria.
*Corresponding author:[email protected]
Article History: Received12October2017; Accepted05February2018 c2018 MJM.
Contents
1 Introduction. . . 331
2 The abstract setting of the problem. . . 332
2.1 Change of variables . . . 332
2.2 Statement of the abstract problem. . . 333
3 Regularity results for the transformed problem. . 334
3.1 Representation of the solution . . . 334
3.2 Study of regularity of the solution . . . 335
4 Regularity results for problem (1.2)-(1.3). . . 336
Acknowledgments. . . 337
References. . . 337
1. Introduction
The regularity analysis for boundary value problems set on sin-gular domains has attracted the attention of many researchers. The solvability of such problems was discussed by means of different methods such as the variational methods or the well known potential theory. For more information, we can refer the reader to [13], [14], [15] and the references cited therein. In this work, we consider the cusp domainΠ⊂R3defined by
Π=
(x,y,z)∈R3: 0<z<1,
x
zα,
y zα
∈D
, α>1,
(1.1)
where
D=
(x,y)∈R2: 0<x2+y2<1 .
We want to study the problem
∂t2u+∆u=h, on]0,1[×Π, (1.2)
associated to the following mixed boundary conditions
∂t2u
{0}×D+∆(x,y)u {
1}×D=u0,
∂t2u
{1}×D−∆(x,y)u {
0}×D=u1, u|]0,1[×∂D=0.
(1.3)
Here,∆stands for the standard Laplacian inR3andu0,u1are
given functions.
We assume that the second memberhis taken in the Lebesgue
spacesLp(0,1;Lp(Π)),3
2 <p<∞. Note here that the bound-ary conditions (1.3) are a considered as restriction of the following ones
∂t2u{
0}×D+∆u|{1}×D=u0, ∂t2u{
1}×D+∆u|{0}×D=u1,
u|]0,1[×∂D=0.
(1.4)
of a forthcoming study. In our situation, the boundary of the domainΠgiven by (1.1) is at the origin of several difficulties which make the use of the classical tools a complicated task. Then, we have opted for the use of another approach based essentially on the theory of abstract differential equations. The effectiveness of this method was manifested in several works, see for example [2], [3], [5], [7], [8], [18] and the references cited therein. Our purpose is to establish existence, uniqueness and maximal regularity of the strict solution for (1.2)-(1.3). For this reason, some suitable compatibility con-ditions onu0andu1will be imposed.
We will prove that the study of our problem can be reduced to the study of an abstract differential equation associated to some nonlocal boundary conditions involving a linear un-bounded operator. The techniques of investigation used here are based on the use of analytic semigroup’s approach com-bined with the sum’s operators theory developed in [6]. This work is organized as follows, in the next section, we show that our problem can be transformed by a natural changes of variables into an abstract second order differential problem. Section 3, is devoted to the study of the abstract version of the transformed problem. Finally, in section 4, we go back to our first problem in the non regular cylindrical domain and we justify our main results.
2. The abstract setting of the problem
2.1 Change of variables
First, we use the following change of variables
T: ]0,1[×Π→]0,1[×Ω,
(t,x,y,z)7→(t,ξ,η,λ), λ>λ0 (2.1)
where
(t,ξ,η,λ) =
t,zxα, y zα,
θ
zβ
Ω=D×]λ0,+∞[,
λ0=α1−1>0,
β =α−1=1
θ.
Putting
v(t,ξ,η,λ) =u(t,x,y,z)andg(t,ξ,η,λ) =h(t,x,y,z).
Then, equation (1.2) is transformed into a new one given by
∂t2v−
θ
λ −2α/β
∆v+
θ
λ −2α/β
Pv=g, (2.2) with
Pv(t,ξ,η,λ)
= (
1+α2θ2
ξ
λ
2)
∂ξ2v+
1+α2θ2
η
λ
2
∂η2v
+2α2θ2ξ η
1
λ 2
∂ξ η2 v+2α θ
ξ
λ
∂ξ λ2 v (2.3)
+2α θ
ξ2
λ
∂η λ2 v+α(α+1)θ2ξ
1
λ 2
∂ξv
+α(α+1)θ2η
1
λ 2
∂ηv
+α θ
1
λ
∂λv.
Now, let us introduce the following change of functions
v= θ λ
s
w,
and
f = ((α−1)λ)
−3α p(α−1)g,
with
s=− α
α−1
3
p−2
=−α
β
3
p−2
.
Consequently, equation (2.2) becomes
k(λ)∂t2w−∆w+1
λLw=f, (2.4)
with
k(λ) =
λ
θ s
,
and
Lw(t,ξ,η,λ)
= (α θ) 2
λ n
ξ2∂ξ2w+η2∂η2w+2ξ η ∂ξ η2 w
o
+2α θ n
ξ ∂ξ λ2 w+ξ2∂η λ2 w o
+ (α θ−2s)∂λw
+α θ
λ ((α+1)θ−2s)
ξ ∂ξw+η ∂ηw
+s
λ(s+1−α θ)w.
Concerning the boundary conditions, it is easy to see that (1.3) are transformed to
k(λ)∂t2w(1,ξ,η,λ)−∆(ξ,η)w(0,ξ,η,λ) =k(λ)w0,
k(λ)∂t2w(0,ξ,η,λ) +∆(ξ,η)w(0,ξ,η,λ) =k(λ)w1,
w(t,ξ,η,λ) =0, (t,ξ,η,λ)∈]0,1[×∂Ω,
2.2 Statement of the abstract problem
Without loss of generality, we consider the following simpli-fied problem
∂t2w−∆w=f, (2.5)
under the corresponding conditions
∂t2w(1,ξ,η,λ)−∆(ξ,η)w(0,ξ,η,λ) =w0, (ξ,η,λ)∈Ω,
∂t2w(0,ξ,η,λ) +∆(ξ,η)w(0,ξ,η,λ) =w1, (ξ,η,λ)∈Ω, w(t,ξ,η,λ) =0, (t,ξ,η,λ)∈]0,1[×∂Ω.
(2.6)
Now, setE=Lp(Ω)andX=Lp(0,1;E)with3
2 <p<∞and consider the following vector-valued functions
w : ]0,1[→E;t−→w(t); w(t) (ξ,η,λ) =w(t,ξ,η,λ),
f : ]0,1[→E;t−→ f(t); f(t) (ξ,η,λ) =f(t,ξ,η,λ).
Set
D(M) =W2,p(Ω)∩W01,p(Ω),
Mψ=∆ψ, ψ∈D(M).
D(N) =W2,p(D)∩W01,p(D),
Nψ=∆(ξ,η)ψ,ψ∈D(N).
and define the following operators.
D(A) ={ψ∈X :ψ(t)∈D(M), a.et∈]0,1[},
(Aψ)(t) =M(ψ(t)), t∈]0,1[,
(2.7)
D(H) ={ψ∈X :ψ(t)∈D(N), a.et∈]0,1[},
(Hψ)(t) =N(ψ(t)), t∈]0,1[.
(2.8)
Consequently, the abstract version of (2.5)-(2.6) is given by
w00(t) +Aw(t) =f(t), t∈]0,1[, (2.9)
w00(1)−Hw(0) =w1, w00(0) +Hw(1) =w0, (2.10)
wherew0,w1are given elements of the complex Banach space
E.Our purpose is to find a strict solution of Problem (2.9)-(2.10), i.e. a functionwsuch that
w∈W2,p(0,1;E)∩Lp(0,1;D(A)),
w0,w1∈D(H),
wsatisfies (2.10).
In the sequel, we need the following results describing some spectral properties of the above cited operators.
Definition 2.1. Let E be a complex Banach space and B be a closed linear operator in E, denoting byρ(B)its resolvent set. Then, B is said to be sectorial if there are constantsµ
∈R,δ ∈(π/2,π), M>0such that
ρ(B)⊃Sδ,µ={z∈C,z6=µ: arg(z−µ)<δ},
(B−zI)−1 L(E)6
C |z−µ|
, z∈Sδ,µ.
(2.11)
Lemma 2.2. The densely closed linear operator(A,D(A))
defined by (2.7) is a sectorial operator withµ=0andδ =π2.
Proof. See section 3.1.1 in [16], in where a complete study of more general elliptic operator was discussed.The density ofD(A)follows from Proposition 2.1.1 in [11].
Similarly, one has
Lemma 2.3. The densely closed linear operator(H,D(H))
defined by (2.8) is a sectorial operator withµ=0andδ =π2.
Remark 2.4. In our situation, observe that
1. it is easy to see that
AH=HA,
2. the estimate (2.11) implies that
Q=−√−A,
is well defined. Furthermore, there exists a sector
Πδ,r0
= {z∈C∗:|argz| ≤δ+π/2} ∪B(0,r0),
(with some positiveδ, r0) and C>0such that
ρ(Q)⊃Πδ,r0
and
∀z∈Πδ,r0,
(Q−zI)−1≤
C |z|.
Thus, one has for all t∈]0,1[andϕ∈E,
etQϕ= 1
2iπ
Z
γ1
etz(Q−zI)−1ϕdz,
whereγ1is a suitable sectorial curve in the complex
plane.
From Remark 1 in [4], it follows that
Lemma 2.5. For anyϕ∈E,k∈N∗and t∈]0,1[, one has
eQϕ∈D Qk,
eQϕ∈Lp(0,1;E),
and
QketQeQϕ∈Lp(0,1;E),
QketQeQϕ=etQQkeQϕ,
3. Regularity results for the transformed
problem
3.1 Representation of the solution
Using the Krein’s method, we know that the unique solution wof (2.9)-(2.10) is given by
w(t)
= b1etQ+b2e(1−t)Q+I(t) +J(t),
where
I(t) = Q
−1
2 t
Z
0
eQ(t−s)f(s)ds,
J(t) = Q
−1
2
1
Z
t
eQ(s−t)f(s)ds.
The constantb1andb2are uniquely determined via the
bound-ary conditions (2.10). For more informations about this tech-nique, we refer the reader to [1] and the references therein. Using the commutativity of the two operators, namely(A,D(A)), (H,D(H)), we obtain the following abstract system
Q2eQ−H
b1+ Q2−HeQb2 =w1−Q2I(1) +HJ(0),
Q2+HeQb1+ Q2eQ+H
b2 =w0−Q2J(0)−HI(1).
(3.1)
The abstract determinant of (3.1) is given by
Λ=− 1−e2Q
Q4+H2,
and one has
Lemma 3.1. The operator
Λ
= − 1−e2Q
Q4+H2 ,
is closed and boundedly invertible. Furthermore
Λ−1 (3.2)
= − Q4+H2−1
1−e2Q−1
, (3.3)
Proof. Thanks to Proposition 2.3.6 in [16], we know that the operator 1−e2Q
has a bounded inverse
1−e2Q−1
= 1
2iπ
Z
γ1
g(z)−1
g(z) (Q−zI)
−1dz+I,
where
g(z) =1−e2z,
andγ1is a suitable curve in the complex plane.
On the other hand, The Dore-Venni sum’s theory allows us to confirm that
Q4+H2,
is closed and Q4+H2−1∈L(E). Furthermore, one has
Q4+H2−1
= − 1
2iπ
Z
γ2
(AH)2zH−2
sinπz dz,
γ2is also a suitable curve in the complex plane, see Theorem
2.1 in [6].
Using the Dunford’s operational calculus, we write
Q4+H2−1
1−e2Q−1
=
1
2iπ
2 Z
γ1×γ2
(AH)2zH−2 sinπz
g(µ)−1
g(µ) (Q−µI)
−1dzd µ
− 1 2iπ
Z
Γ2
(AH)2zH−2 sinπz dz
At this level, Fubini’s theorem allows us to write
Q4+H2−1
1−e2Q−1 =
−
1 2iπ
Z
γ1
g(µ)−1
g(µ) (Q−µI)
−1d µ
(×)
−
1 2iπ
Z
γ2
(AH)2zH−2 sinπz dz
− 1 2iπ
Z
γ2
(AH)2zH−2 sinπz dz
= 1−e2Q−1
Q4+H2−1 .
Summing up, we deduce that the formal solution of (2.9)-(2.10) is given by the formula
w(t)
= etQΛ−1Hw1−Q2w0
+etQeQΛ−1Hw0+Q2w1
+e(1−t)QΛ−1Hw0+Q2w1 (3.4)
−e(1−t)QeQ
Λ−1Hw1−Q2w0
+etQ 1−e2Q−1J(0)
+etQeQ
h
2Λ−1Q4I(1) + 1−e2Q−1I(1)i
+e(1−t)Q 1−e2Q−1[I(1)−J(0)]
Putting
d0=Λ−1
Hw1−Q2w0
,
d1=Λ−1Hw0+Q2w1
,
¯
d0= 1−e2Q
−1
J(0),
¯
d1=2Λ−1Q4I(1) + T I(1),
¯
d2= 1−e2Q −1
[I(1)−J(0)].
(3.5)
Then,wis given by the following compact expression
w(t) =G1(t) +G2(t) +G3(t) +G4(t),
with
G1(t) =etQd0+e(1−t)Qd1+etQd¯0,
G2(t) =etQeQd1−e(1−t)QeQd0,
G3(t) =etQd¯0+etQeQd¯1+e(1−t)Qd¯2,
G4(t) =I(t) +J(t).
3.2 Study of regularity of the solution From [6]), one has
Lemma 3.2. . Let f ∈Lp(0,1;E), 32<p<∞. Then, the following applications
t→Q t
R
0
e(t−s)Qf(s)ds,
t→Q
1
R
t
e(s−t)Qf(s)ds,
are well defined for a.e. t∈]0,1[and belong to Lp(0,1;E).
To establish more regularity results, we need to introduce for any closed operatorBandδ ∈]0,1[,the following
clas-sical real interpolation spaces betweenD(B)andEdefined by
(E,D(B))δ,p= (D(B),E)1−δ,p
For more details about these spaces, see [9]. In our situation, sinceQgenerates an analytic semigroup, it follows from [17] that beginlemma For all32<p<∞andϕ∈E, one has
Qe.Qϕ
∈Lp(0,1;E)⇔ϕ∈(D(A),E)1 2p+12,p,
and Ae.Qϕ
∈Lp(0,1;E)⇔ϕ∈(D(A),E)1 2p,p
.
Keeping in mind that I(0)and J(1)are bounded and due to the regularizing effect of eQ, Λ−1and T,the following
regularity results are a direct consequence of Lemma3.2.
Proposition 3.3. Let d0, d1,d¯0,d¯1,d¯2given by (3.5). Then,
for 32<p<∞, one has
1. t → Q2G1(t) ∈ Lp(0,1;E) if and only if
d0, d1∈ D(Q2),E
1 2p,p
.
2. t→Q2G2(t)∈Lp(0,1;E).
3. t→Q2G3(t)∈Lp(0,1;E).
4. t→Q2G4(t)∈Lp(0,1;E).
The above proposition can be equivalently formulated as follows
Theorem 3.4. Let f∈Lp(0,1;E), 32<p<∞. and w0, w1∈
E Then, the following assertions are equivalents
1. Hw1,Aw0, Hw0,Aw1∈(D(A),E)1 2p,p
.
2. w given by (3.4) is the unique strict solution of (2.9 )-(2.10) satisfying the maximal regularity property, that is
w00and Aw∈Lp(0,1;E).
In order to state our main results, we recall the definition of the following Besov space forυ∈(0,1)
Bυ
p,p(Ω)
: =
ψ∈Lp(Ω):
Z
Ω Z
Ω
|ψ(ξ1)−ψ(ξ2)|p
kξ1−ξ2k1+pυ
dξ1dξ2<∞
whereξ1andξ2are a two generic points ofR3, see [9], p.
680, and one has
Lemma 3.5. Let A the operator defined by (2.7) and 32<p<
∞. Then
(D(A),E)1 2p+
1 2,p
=
ψ∈Lp 0,1;Bυp(Ω): ψ=0on∂Ω
: =Lp0,1; ˚Bυ
p(Ω)
.
Proof. One has
(D(A),E)1 2p+12,p
=Lp
0,1;W02,p(Ω)∩Lp(Ω)
1 2p+12,p
First, it is well known that
Lp
0,1;W02,p(Ω)
⊂ Lp
0,1;W02,p(Ω)∩Lp(Ω)
1 2p+
1 2,p
⊂ Lp(0,1;Lp(Ω)).
From [9], Proposition 3, P. 683, one has
W2,p(Ω)∩Lp(Ω)1 2p+12,p
=B1−
1
and as in p. 708 in the same work cited above, we get
W02,p(Ω)∩Lp(Ω)
1 2p+
1 2,p
=
ψ∈B 1−1
p
p (Ω):ψ=0 on∂Ω
: =B˚1−
1
p p (Ω)
Then, we can formulate our main results in the trans-formed domain as follows.
Theorem 3.6. Let f∈Lp((0,1)×Ω),32<p<∞. Then, the following assertions are equivalents
1. ∆(ξ,η)w1,∆w0,∆(ξ,η)w0,∆w1∈Lp
0,1; ˚B1−
1
p p (Ω)
.
2. Problem (2.5)-(2.6) has a unique strict solution
w∈W2,p(0,1;Lp(Ω))∩Lp(0,1;W02,p(Ω)),
satisfying the maximal regularity property, that is
∂t2w and∆w∈LP(0,1;Lp(Ω)).
Using identically a classical perturbation argument as in [10], we conclude that
Theorem 3.7. Let f ∈Lp((0,1)×Ω), 32<p<∞. Then the following assertions are equivalents.
1. ∆(ξ,η)w1,∆w0,∆(ξ,η)w0,∆w1∈Lp
0,1; ˚B1−
1
p p (Π)
.
2. Problem (2.4)-(2.6) has a unique strict solution
w
∈ W2,p(0,1;Lp(Ω))∩Lp(0,1;W02,p(Ω)),
satisfying the maximal regularity property, that is
∂t2w and∆w∈Lp(0,1;Lp(Ω)).
4. Regularity results for problem
(1.2)-(1.3)
From [2], the go back to our original domain use the following inverse change of variables
T−1: ]0,1[×Ω→]0,1[×Π,
(t,ξ,η,λ)7→
t, θ λ
α/β
ξ, θλα/βη, θλ1/β
.
(4.1)
Since
w =
θ
λ 3α/pβ
z−2αu
= z3α/pz−2αu,
then, a direct computation leads to
(
∂ξw= θ λ
3α/pβ
z−α∂
xu=z3α/pz−α∂xu,
∂ηw= θλ
3α/pβ
z−α∂
yu=z3α/pz−α∂yu,
and
∂ξ2w= θλ3α/pβ∂x2u=z3α/p∂x2u,
∂η2w= θ λ
3α/pβ
∂y2u=z3α/p∂2
yu,
∂ξ η2 w= θλ3α/pβ∂xy2u=z3α/p∂xy2u.
As an immediate consequence of the preceding maximal reg-ularity results with respect to(t,ξ,η,λ), we deduce then that
z−2αu,z−α
∂xu,z−α
∂yu,∂xy2u,∂x2u,∂y2u
∈ Lp(0,1;Lp(Π)), 3
2 <p<∞
On the other hand, since∂λ2w=∆w−∆(ξ,η)w, then
∂z2u∈Lp(0,1;Lp(Π)).
Regarding now the cross derivatives∂ξ λ2 wand∂η λ2 w, one has
∂ξ λ2 w=
θ λ
3α/pβ
α
2α−3α
p
θ λz
−α
∂xu
−αθλ ∂x2u+∂xy2u−∂xz2u
,
and
θ λ
3α/pβ
α
2α−3pα
θ λz
−α∂yu
−αθλ ∂y2u+∂xy2u−∂xz2u
,
from which, we deduce that
∂xzu,∂yzu∈Lp(0,1;Lp(Π)), 3
2<p<∞.
Finally, one has
∂λw =
θ
λ
3α/pβ
2α−3α
p
θ
λz
−2αu
−
θ
λ
3α/pβ αθ
λz
−α(∂
xu+∂yu)−z−α∂zu
,
it follows then that
z−α
∂zu∈Lp(0,1;Lp(Π)), 3
2 <p<∞
Summing up, one has
z−2αu,z−α
∂xu,z−α∂yu,z−α∂zu,
Similarly, the concrete version of the compatibility conditions onu0andu1are given by
z−2αu
0,z−α∂xu0,z−α∂yu0,z−α∂zu0,z∂xy2u0,∂x2u0,∂y2u0
∈B˚2−
1
p,p p,# (Π),
and z−2αu
1,z−α∂xu1,z−α∂yu1,z−α∂zu1,∂xy2u1,∂x2u1,∂y2u1
∈B˚2−1p,p p,# (Π).
where
˚ B2−1p,p
p,# (Π):=
ψ:z3αpψ∈B˚2−
1
p,p p (Π)
.
Summing up, we are in position to state our main results
Theorem 4.1. Let h∈Lp(0,1;Lp(Π)), 32<p<∞.Assume that
z−2αu
0,z−α∂xu0,z−α∂yu0,z−α∂zu0,z∂xy2u0,∂x2u0,∂y2u0
∈B˚2−1p,p p,# (Π),
and z−2αu
1,z−α∂xu1,z−α∂yu1,z−α∂zu1,∂xy2u1,∂x2u1,∂y2u1
∈B˚2−1p,p p,# (Π).
Then problem (1.2)-(1.3) has unique strict solution
u∈W2,p(0,1;Lp(Π))∩Lp(0,1;W02,p(Π)),
such that
∂t2u
∈Lp(0,1;Lp(Π)),
and z−2αu,z−α
∂xu,z−α∂yu,z−α∂zu,∂xy2u,∂xzu,∂yzu,∂x2u,∂y2u,∂z2u ∈Lp(0,1;Lp(Π)).
Acknowledgments
The authors would like to express sincere gratitude to the reviewers for his/her valuable suggestions.
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